Consider a triangle ABC. Let the angle bisector of angle A intersect side BC at a point D between B and C. The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC:

{\displaystyle {\frac {|BD|}{|DC|}}={\frac {|AB|}{|AC|}},} {{\frac {|BD|}{|DC|}}}={{\frac {|AB|}{|AC|}}},The vector $A,B,C$ of triangle $ABC$ have respectively position vectors $\vec{a}, \vec{b}, \vec{c}$ with respect to a given origin $O$. Show that the point $D$ where the bisector of $\angle A$ meets $BC$ has position vector $\vec{d}=\frac{\beta \vec b+\gamma \vec c}{\beta+\gamma}$, where $\beta=|\vec c-\vec a|$ and $\gamma=|\vec a-\vec b|$. Hence deduce that the incentre $I$ has position vector $\frac{\alpha \vec a+\beta \vec b+\gamma \vec c}{\alpha+\beta+\gamma}$, where $\alpha=|\vec b-\vec c|$


We can prove this using the result "The angle of bisector of the angle $\angle A$" of a triangle $ABC$ divides the sides $BC$ in the ration $AB:AC$.

Is there any other way to prove this result?


Since the angle bisector of $\widehat{BAC}$ is the locus of points $P$ for which $d(P,AB)=d(P,AC)$, the trilinear coordinates of the incenter $I$ are $[1;1;1]$. It follows that the barycentric coordinates of the incenter are $[a;b;c]$, hence the exact barycentric coordinates are $\left[\frac{a}{a+b+c};\frac{b}{a+b+c};\frac{c}{a+b+c}\right]$, i.e.

$$ I = \frac{aA+bB+cC}{a+b+c} $$ as wanted.


A simple (although possibly tedious) approach is to prove it algebraically. I will simply write the lowercase letters for the vectors.

From the angle equality

$$ \frac{(d-a)\cdot (c-a)}{|c-a|}=\frac{(b-a) \cdot (d-a)}{|b-a|}$$

Since $ d-c$ and $ b-d$ are collinear,

$$ d-c = \delta(b-d) $$ $$ d = \frac{\delta b+ c}{1+\delta}$$ (this is a general result for a point lying on a line determined by two position vectors)

$$ d-a = \frac{\delta (b-a)+(c-a)}{1+\delta}$$

$$ \frac{(d-a).(c-a)}{|c-a|} = \frac{\delta (b-a)\cdot (c-a)}{(1+\delta)|c-a|}+\frac{|c-a|}{1+\delta}$$

$$ \frac{(b-a).(d-a)}{|b-a|} = \frac{\delta |b-a|}{(1+\delta)}+\frac{(c-a)\cdot (b-a)}{(1+\delta)|b-a|}$$

Equate and solve for $\delta$ to get the first part.

For the second part, we use a similar approach. For any two angle bisectors, the incentre will have the general form:

$$ \frac{a+\mu d}{1+\mu} = \frac{b+\eta e}{1+\eta}$$

We get two equations to solve for two variables. This will give the result.


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